G.P 1

The sum of the 1st five terms of a geometrical progression is 484. If the common ratio is 3 find the 4th term.

solution

S₅ = 484, G₁ = ?, r=3, n=5, A₂=?

We use the summation formula to find the 1^st term.

S_n = G₁(rⁿ-1)/r-1

               

S₅ = G₁(r⁵-1)/r-1

               

484= G₁(3⁵-1)/3-1

               

484= G₁(243-1)

                  2

2 x 484= G₁(242)

                  

968= 242 G₁

⁴ 968   =  G₁

  242

G₁ = 4

Now we solve for the 2^nd term.

G_n = G₁r^n-1

G₄ = G₁r^4-1

G₄= G₁r ³

G₄ = 4 x (3)³

     = 4 x 27

     = 81

Hence the 4th term is 81.

TRY THIS.........................

The sum of the 1st five terms of a geometrical progression is 484. If the common ratio is 3 find the 3rd term.

SETS 2

If n(A)= 90 , n(AuB) = 140 and n(AnB)=30, find n(B)

Solution

n(AuB) = n(A) + n(B) - n(AnB)

  140     = 90 + n(B)  – 30

  140     = 90 -30 + n(B) 

  140     = 60 + n(B) 

  140 - 60        = n(B)

  80        = n(B)

Hence n(B) = 80 answer

TRY THIS…………………………….

If n(A)= 70 , n(AuB) = 137 and n(AnB) = 35, find n(B).

LOGARITHMS 3

Evaluate Log100,000,000 +  log 0.00001 + log₃243

Solution

= Log100,000,000  +  log 0.00001 + log₃243

= Log10⁸ +  log 10^-5 + log₃3⁵

= 8Log10 +  (-5log 10) + (5log₃3)    [since log_aaⁿ  = nlog_aa]

= (8x1) + (-5x1) + (5x1)             [since log_aa = 1]

= 8 + (-5) + (5)

= 8

Hence Log100,000,000 +  log 0.00001 + log₃243 = 7

TRY THIS...............................

Evaluate Log10,000,000,000 +  log 0.01 + log₃2187

POLYGONS 1

An interior angle of a regular polygon is 68⁰ greater than an exterior angle. Find the interior angle.

Solution

Let i = interior angle, e = exterior angle.

Now i  + e=180⁰…………………(1)

But i = e+68⁰ …………………(2)

Substitute (2) in (1) above.

e+68⁰   + e=180⁰

e+ e+68⁰   =180⁰

2e+ 68⁰   =180⁰

2e=180⁰ - 68⁰   

2e=112⁰

2e=112⁰          dividing by 2 both sides.

2      2

e = 56⁰

But i  + e=180⁰…………………(1)

i  + 56⁰=180⁰.

i  =180⁰ - 56⁰

i = 124⁰

Hence i = 124⁰ 

TRY THIS………………………   

An interior angle of a regular polygon is 84⁰ greater than an exterior angle. Find the interior angle.

ALGEBRA 1

48 + 24 =56

        a

Solution

48 + 88 =56

        a

   88  = 56 - 48

   a

   88  =  8

    a

¹ a  x  88  =  8 x a

        ₁ a

88 = 8a

88  = 8a

8       8

11 = a

Hence a=11.

TRY THIS.........................

22 + 72 =30    ; find c.

         c

FACTORIZE 1

Factorize 81-9m²

Solution

We use difference of two squares a² – b² = (a - b)(a + b)

81-9m² = 9² - 3²m²       

             = 9² - (3m)²     

             = (9 - 3m)(9 + 3m)

Hence 81 - 9m² = (9 - 3m)(9 + 3m)

TRY THIS…………….

Factorize 225 - 16c²

LOGARITHMS 2

If Log(30x-90/5+x) = 1. Find x.

Solution

Log(30x-90/5+x) = 1.

Log₁₀(30x-90/5+x) = 1.

 30x-90 = 10¹. After changing into exponential form

   5+x

30x-90 = 10    since [10¹=10]

   5+x

30x-90 = 10(5+x)    after cross multiply

  

30x-90 = 50+10x

30x-10x = 50+90 [collecting like terms]

20x = 140

x=7  [after dividing by 20 both sides]

Hence x=7.

TRY THIS……………

If Log(30x-50/5+x) = 1. Find x.

SETS 1

If n(A)= 85 , n(B)= 96 and n(AuB)= 140, find n(AnB).

Solution

n(AuB) = n(A) + n(B) - n(AnB)

140 = 85 + 96 - n(AnB)

140 = 181 - n(AnB)

140 - 181 = - n(AnB)

-41 = -n(AnB)

n(AnB) = 41 [after dividing by -1 both sides]

Hence n(AnB) = 41 answer

TRY THIS………….

If n(A)= 90 , n(B)= 96 and n(AuB)= 139, find n(AnB).

LOGARITHMS 1

  Evaluate Log60 – Log0.3 + Log500.

Solution

= Log60 – Log0.03 + Log500.

= Log(60 x 500)

               0.03

= Log(30000)

             0.03

= Log(3000000)

              3

= Log 1,000,000

             

= Log₁₀1,000,000

= Log₁₀10⁶

            

= 6Log₁₀10

= 6 x 1

= 6

∴ Log600 – Log0.03 + Log50 = 6

TRY THIS……………………….

Evaluate Log6,000 – Log4.8 + Log80,0000

SIMPLIFY 1

Simplify 21x-3(x-y)+5

Solution

=21x-3(x-y)+5

=21x-3x+3y+5

=18x+3y+5 answer

TRY THIS……..

Simplify 44x-9(x-y)+17